How to calculate the perimeter and area of a polygon?

How would I calculate the perimeter and area of any shape given its coordinates, but without using any built in functions in MATLAB? So far this is what I have:

Perimeter

points = [1 0 ; 3 0 ; 3 2 ; 5 2 ; 5 4 ; 3 4 ; 3 6 ; 1 6 ; 1 4 ; -1 4 ; -1 2 ; 1 2 ;];

perimeter = 0;

for i = 1:size(points, 1)-1

perimeter = perimeter + norm(points(i, :) - points(i+1, :));

end

perimeter = perimeter + norm(points(end, :) - points(1, :)); % Last point to first

fprintf('The perimeter of the polygon is %.3f\n', perimeter)

(points is the set of coordinates)

Area

function p_area = assignment_task_b(x,y)

% Get the number of vertices

n = length(x);

% Initialize the area

p_area = 0;

% Apply the formula

for i = 1 : n-1

    p_area = p_area + (x(i) + x(i+1)) * (y(i) - y(i+1));

end

p_area = abs(p_area)/2;

Any feedback and corrections are appreciated.

Answer 1

[Roger Stafford]

The perimeter looks right, but to make the area right either the first point must equal the last point or you need to extend the p_area by one more term, namely, (x(n)+x(1))*(y(n)-y(1)). Also you need to either ensure that your route around the polygon is clockwise or else you should take the absolute value of the result.

You should note that it is possible to avoid the for-loops here with a vectorized expression for both the perimeter and the area.

[Jason Flint]

area = 1/2*sum(x.*y([2:end,1])-y.*x([2:end,1]));

Is this what you mean by a vectorised expression?

Answer 2

[Roger Stafford]

Yes, that is what is meant by a "vectorized" expression. However the sign of your area is now the opposite of the sign of your original expression and would be correct only for going counterclockwise. You can take the absolute value to be independent of which way the path goes.

Answer 3

[Andrei Bobrov]

one way

p1 = points(randperm(size(points,1)),:); % Let your data

ii = bsxfun(@minus,p1,mean(p1))*[1;1i];
[~,jj] = sort(angle(ii));
i1 = [ii(jj);ii(jj(1))];
a1 = mod(diff(angle(i1)),2*pi);
v = abs(i2);

area1 = sum(prod([v(1:end-1),v(2:end),sin(a1)]));
perim = sum(abs(diff(i2)));